Lamb’s quarters is a common weed that interferes with the growth of corn. An agriculture researcher planted corn at the same rate in $16$small plots of ground and then weeded the plots by hand to allow a fixed number of lamb’s quarters plants to grow in each meter of cornrow. The decision on how many of these plants to leave in each plot was made at random. No other weeds were allowed to grow. Here are the yields of corn (bushels per acre) in each of the plots:

Here is some computer output from a least-squares regression analysis of these data. Do these data provide convincing evidence at the $\mathrm{\alpha }=0.05$level that more lamb’s quarters reduce corn yield?

$$\begin{gathered} \mathrm{Predictor}\mathrm{Coef}\mathrm{SE}\mathrm{Coef}\mathrm{T}\mathrm{P} \\ \mathrm{Constant}166.4832.72561.110.000 \\ \mathrm{Weeds}\mathrm{per}-1.09870.5712-1.920.075 \\ \mathrm{meter} \\ \mathrm{S}=7.97665\mathrm{R}-\mathrm{Sq}=20.9\%\mathrm{R}-\mathrm{Sq}\left(\mathrm{adj}\right)=15.3\% \end{gathered}$$

We need to find the given data which provides convincing evidence at the $\mathrm{\alpha }=0.05$level that more lamb’s quarters reduce corn yield.

Consider:

$$\begin{gathered} \mathrm{n}=\mathrm{Sample}\mathrm{size}=16 \\ \mathrm{\alpha }=\mathrm{Significance}\mathrm{level}=0.05 \end{gathered}$$

The estimate of the slope ${\mathrm{b}}_1$is given in the row "Weeds per meter" and in the column "Coef" of the given computer output:

${\mathrm{b}}_1=-1.0987$

The estimated standard deviation of the slope ${\mathrm{SE}}_{\mathrm{b}1}$is given in the row "Weeds per meter" and in the column "SE Coef" of the given computer output:

${\mathrm{SE}}_{\mathrm{b}1}=0.5712$

Given claim: Slope is negative (reduction):

The null hypothesis or the alternative hypothesis states the given claim The null hypothesis states that the slope is zero. If the given claim is the null hypothesis, then the alternative hypothesis states the opposite of the null hypothesis.

role="math" localid="1654162502097" $$\begin{gathered} {\mathrm{H}}_0:{\mathrm{\beta }}_1=0 \\ {\mathrm{H}}_{\alpha }:{\mathrm{\beta }}_1<0 \end{gathered}$$

Compute the value of the test statistic:

$\mathrm{t}=\frac{{\mathrm{b}}_1-{\mathrm{\beta }}_1}{{\mathrm{SE}}_{{\mathrm{b}}_1}}=\frac{-1.0987-0}{0.5712}\approx -1.9235$

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of the Student's T table in the appendix containing the -value in the row $\mathrm{df}=\mathrm{n}-2=16-2=14$We can ignore the minus sign in the test statistic:

$0.025<\mathrm{P}<0.05$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:
$\mathrm{P}<0.05\Rightarrow \mathrm{Reject}{\mathrm{H}}_0$